Should we use relativistic corrections for particles?

[pattiecake]

Does anyone know if we should use relativistic corrections for particles? Say an electron accelerating out of an electron gun at 2000V. How would you calculate the final velocity?

 

[da_willem]

Does anyone know if we should use relativistic corrections for particles? Say an electron accelerating out of an electron gun at 2000V. How would you calculate the final velocity?

Classically the change in kinetic energy $$m_ev^2$$ will equal the change in potential energy $$q_e V$$. This yields $$v=\sqrt{\frac{q_e V}{m_e}$$. With V=2000V, v=0,06256c. If you're only interested in a rough estimate you're done, but a refinement can be made treating the problem relativistically.

Relativistically conservation of energy involves the potential energy $$q_e V$$ and total relativistic energy: $$\frac{m_e c^2}{\sqrt{1-v^2/c^2}}$$. Letting the sum of these energies be the same before and after the acceleration of the electron yields:

$$q_e V+m_e c^2=0+\frac{m_e c^2}{\sqrt{1-v^2/c^2}}$$
$$\frac{q_eV}{m_e c^2}=1-\frac{1}{\sqrt{1-v^2/c^}}$$

Which you can solve for $$v$$. Often it is easier to use the relationship between relativistic momentum $$p=\frac{mv}{\sqrt{1-v^2/c^2}}$$ and energy:

$$E^2=c^2p^2 + m^2c^4$$

With $$E=q_e V+m_e c^2$$ this makes:

$$\sqrt{\frac{(q_e V+m_e c^2)^2}{c^2} -(m_e)^2 c^2} = p = \frac{m_e v}{\sqrt{1-v^2/c^2}}$$

Filling in numbers you get a final velocity of v=0,088216c

 

[pattiecake]

neat! Thanks!

 

[turin]

da willem,
did you get a HIGHER velocity with the relativistic correction?

 

[da_willem]

ehm, yes. This can't be right... I'll check it later...

 

[pmb_phy]

Does anyone know if we should use relativistic corrections for particles?

As da_willem said, yes. In the present case conservation of energy is given by

Kinetic energy + potential energy = K + V = constant

When this equation is derived it is assumed that the proper mass (aka "rest mass") of the particle is constant. Let m_e be the proper mass of the electron (which obviously doesn't change). This gives, upon substitution of $$K = (\gamma - 1)m_ec^2$$

$$K + V = (\gamma - 1)m_ec^2 + V = constant$$

where V is the potential energy (I use "V" to mean something different than da_willem).

This can be written as

$$\gamma_{1} m_ec^2 - m_ec^2 + V(\mathbf{r}_1) = \gamma_{2} m_ec^2 - m_ec^2 + V(\mathbf{r}_2)$$

Canceling the common factor gives

$$\gamma_{1} m_ec^2 + V(\mathbf{r}_1) = \gamma_{2} m_ec^2 + V(\mathbf{r}_2) = constant$$

The potential energy V is related to the Coulomb potential $$\Phi$$ as

$$V = q_e\Phi$$

Thus

$$\gamma_{1} m_ec^2 + q_e\Phi(\mathbf{r}_1) = \gamma_{2} m_ec^2 + q_e\Phi(\mathbf{r}_2)$$

Another way to simplify is to define the quantity E = K + E₀. Then

$$E = K + E_0 = (\gamma - 1)m_ec^2 + m_ec^2 = \gamma m_ec^2 = mc^2$$

where $$m = \gamma m_ec^2$$ is the (relativistic) mass of the electron. This gives

$$E + V = mc^2 + V = mc^2 + q\Phi = \gamma m_ec^2 + q_e\Phi = constant$$

as given above. The Lagrangian formulation may also be used to obtain a constant of motion known as Jacobi's integral since its an "integral of motion", aka a constant of motion. For the derivation see

http://www.geocities.com/physics_world/sr/relativistic_energy.htm

What value you get for Jacobi's integral depends on an arbitrary constant that you add to the Lagrangian.

See also -- http://www.geocities.com/physics_world/sr/work_energy.htm